We will have to do the same thing for the parallel lines, but not with both the directional vectors, since they go the same direction. What I did is using a point from both lines and subtract one from the other, obtaining a vector that is not parallel to them but that we will need to get the plane. We proceed the same way as before, using a point of one of the two lines. I then find myself with ##\pi : 2x - 8y + 5z - 7 = 0##

Now it comes my problem: the distance between the two parallel lines.
I know that you just have to find the distance between one of the lines and a point of the other line.
Let's start from the line ##t##. The directional vector and a point are respectively ##\vec v_t = (-12, -4, 8)## and ##P_t (-1, -1, 1)##
Now I start by finding a plane orthogonal to the ##r##(with ##\vec v_r = (-3, -1, 2)##) line passing for a point belonging to it ##P_r (0, 1, 3)##.
Similar to before, I would have ##\pi : -3x - y + 2z - 5##
Now I have to find the projection of the point ##P_r## on the line ##t##(for example calling it ##P'##) as an intersection between the line and the plane.
##\begin{cases}
-3x - y - 2z - 5 = 0 \\
4x - 2y + 5z - 3 = 0 \\
2y + z + 1 = 0
\end{cases}
\begin{cases}
x = -\frac{5}{3} \\
y = -\frac{11}{9} \\
z = \frac{13}{9}
\end{cases}##
Getting the point ##P' (-\frac{5}{3}, -\frac{11}{9}, \frac{13}{9})##
Now I simply have to find the distance between ##P_r## and ##P'## as:
##d(P_r, P') = \sqrt{(0 - (-\frac{5}{3}))^2 + (1 - (-\frac{11}{9}))^2 + (3 - \frac{13}{9})^2}##
The problem is that I end up with something like this ##\sqrt{\frac{821}{81}}## which is just odd and I don't know if I did it correctly. Could someone tell me if my way of resolving it is correct at least?